What is (\partial)/(\partial x)((e^{xz+y})/(z^2+w)) ?

The answer to (\partial)/(\partial x)((e^{xz+y})/(z^2+w)) is (e^{xz+y}z)/(z^2+w)

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(\partial)/(\partial x)((e^{xz+y})/(z^2+w))

\frac{\partial}{\partial x} (\frac{e ^{x z + y}}{z ^{2} + w})

\frac{e ^{x z + y} z}{z ^{2} + w}

Solution steps

\frac{\partial}{\partial x} (\frac{e ^{x z + y}}{z ^{2} + w})

Treat

z, y, w

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{z ^{2} + w} \frac{\partial}{\partial x} (e^{x z + y})

Apply the chain rule:

e^{x z + y} \frac{\partial}{\partial x} (x z + y)

= e^{x z + y} \frac{\partial}{\partial x} (x z + y)

\frac{\partial}{\partial x} (x z + y) = z

= \frac{1}{z ^{2} + w} e^{x z + y} z

Simplify

\frac{1}{z ^{2} + w} e^{x z + y} z : \frac{e ^{x z + y} z}{z ^{2} + w}

= \frac{e ^{x z + y} z}{z ^{2} + w}

\frac{d}{d x} (\frac{1}{( x + 9 ) ^{2}})

\frac{\partial}{\partial x} ((- x^{4} + 7 y^{2} + 3 y)^{6})

\int 1 + \frac{1}{2 x} d x

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 2

\int_{1}^{3} e^{x} d x

What is (\partial)/(\partial x)((e^{xz+y})/(z^2+w)) ?
The answer to (\partial)/(\partial x)((e^{xz+y})/(z^2+w)) is (e^{xz+y}z)/(z^2+w)